Answer:
1. 0/6 (7 is not on a standard die!)
2. 1/6 (2 is ONE of the numbers on a die so ONE out of 6)
3. 1/6(1 is ONE of the number on a die)
4. 5/6 (there are 5 other numbers on a die other than 1 so 5/6)
5. 2/6 (there are two numbers that start with f, four and five.)
Answer:
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Step-by-step explanation:
Using Pythagoras Theorem:
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Answer:
Step-by-step explanation:
Letting "a" and "c" represent the costs of adult tickets and child tickets, the problem statement gives us two relations:
3a +5c = 52
2a +4c = 38
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We can solve this system of equations using "elimination" as follows:
Dividing the first equation by 2 we get
a +2c = 19
Multiplying this by 3 and subtracting the first equation eliminates the "a" variable and tells us the price of a child ticket:
3(a +2c) -(3a +4c) = 3(19) -(52)
c = 5 . . . . . . collect terms
a +2·5 = 19 . . substitute for c in the 3rd equation above
a = 9 . . . . . . subtract 10
One adult ticket costs $9; one child ticket costs $5.
Answer:
Step-by-step explanation:
Let the number of cockles be c and that of winkles be w.
From the first relation,
w = 2c ..........(I)
From the second relation, 6 winkles are accidentally split.
This in equation form means the following:
c/w- 6 = 3/5
By cross multiplication,
5c = 3(w - 6)...........(ii)
We know w = 2c , let’s substitute this into the second equation.
5c = 3(2c - 6)
5c = 6c - 18
Collecting like times, c = 18
And w = 2c = 2(18) = 36
Hence, there were initially 18 cockles and 36 winkles
